On a divergent integral

  • Thread starter mmzaj
  • Start date
  • #1
107
0

Main Question or Discussion Point

greetings . we have the integral :
[tex] \lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds[/tex]

which diverges for every value of n except [itex] n=0 [/itex]
if we perform the change of variables :

[tex] s\rightarrow \frac{1}{s}[/tex]

then :
[tex]\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds=\int_{-i}^{i}\frac{(1-s)^{n}}{s^{n+1}}ds[/tex]

which converges . am i missing something here , or is this correct !?
 

Answers and Replies

  • #2
1,796
53
greetings . we have the integral :
[tex] \lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds[/tex]

which diverges for every value of n except [itex] n=0 [/itex]
if we perform the change of variables :
Can you solve it for n=2 using antiderivatives? That is, what is:

[tex]\lim_{T\to\infty}\left(-2s+\frac{s^2}{2}+\log(s)\right)_{2-iT}^{2+iT}[/tex]
 
  • #3
mathman
Science Advisor
7,829
433
greetings . we have the integral :
[tex] \lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds[/tex]

which diverges for every value of n except [itex] n=0 [/itex]
if we perform the change of variables :

[tex] s\rightarrow \frac{1}{s}[/tex]

then :
[tex]\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds=\int_{-i}^{i}\frac{(1-s)^{n}}{s^{n+1}}ds[/tex]

which converges . am i missing something here , or is this correct !?
You need to split the original integration range into two parts at s = 0. Now when you change s to 1/s, you will be able to get the correct integration limits. Note also that the original T becomes 1/T.
 

Related Threads on On a divergent integral

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
20
Views
5K
Replies
1
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
4
Views
2K
Top